If $\int {\tan }^{7}xdx=f\left(x\right)+\log \left|\cos x\right|+c$ then $f\left(x\right)=$

We know that  $\int {\tan }^{n}xdx=\frac{{\tan }^{n-1}x}{n-1}-I_{n-2}$

$\int {\tan }^{7}xdx$

$=\frac{{\tan }^{6}x}{6}-I_5$

$=\frac{{\tan }^{6}x}{6}-\left(\frac{{\displaystyle {\tan }^{4}x}}{{\displaystyle 4}}-I_3\right)$

$=\frac{{\tan }^{6}x}{6}-\frac{{\tan }^{4}x}{4}+\frac{{\tan }^{2}x}{2}-\int \tan x+c$

$=\frac{{\tan }^{6}x}{6}-\frac{{\tan }^{4}x}{4}+\frac{{\tan }^{2}x}{2}+\log \left|\cos x\right|+c$

$\therefore f\left(x\right)=\frac{{\tan }^{6}x}{6}-\frac{{\tan }^{4}x}{4}+\frac{{\tan }^{2}x}{2}$